Understanding Inverse Trig Functions: Solving for Phi in Cos Using Inverse Sin

[chrisa88]

How does this work? I'm very confused about the phi is solved using inverse sin.
knowing: A=(c^{2}_{1}+c^{2}_{2})^{1/2} and c_{2}= Acos(\phi)
solve for \phi
which yields: \phi=sin^{-1}\frac{c_{2}}{(c^{2}_{1}+c^{2}_{2})^{1/2}}=tan^{-1}\frac{c_{2}}{c_{1}}
I'm not sure how we use the inverse sin to find the phi in the cos function.
I thought to get the inside of the parenthesis of cos you would use inverse cos, or cos^{-1}. Where am I going wrong?

 

[UltrafastPED]

Your math expressions are yielding an error here.

Just as there are many trigonometric relationships, so there are apparently just as many relations between their inverses. See http://en.wikipedia.org/wiki/Inverse_trigonometric_functions

As always, start with the easiest, defining relationships and build out from there. Note that as sine and cosine are related by complementary angles, so are their inverses.

 

[chrisa88]

I thought this was an error, but the solutions manual to my quantum mechanics class AND the handwritten solutions provided by my professor both have this error. Thank you for confirming!